Answer:
112.8 g.
Explanation:
What is given?
We have water vapor in the closed container:
Q (heat) = 7,500 J.
c (specific heat of water vapor) = 1.9 J/g°C.
ΔT (change of temperature) = Final temperature - Initial temperature = 140 °C - 105 °C = 35 °C.
Step-by-step solution:
To solve this problem, we have to use the formula of specific heat:
[tex]c=\frac{Q}{m\cdot\Delta T}.[/tex]
Where c is the specific heat, in this case, the specific heat of water vapor in J/g°C; m is the mass in g; and ΔT is the change of temperature in °C.
We want to find the mass of the vapor in the container, so we have to solve for 'm' and replace the given data, like this:
[tex]m=\frac{Q}{c\cdot\Delta T}=\frac{7,500\text{ J}}{1.9\frac{J}{g\degree C}\cdot35\degree C}=112.78\text{ g}\approx112.8\text{ g.}[/tex]
The answer would be that the mass of the vapor in the container is 112.8 g.
